Ethanol, with the formula C2H5OH, is also referred to as Ethyl alcohol.
Explanation:
- The interaction between oil and sodium hydroxide (NaOH) is recognized as the Saponification process.
- A combination of ethanol and water yields a fully homogeneous solution, miscible in all proportions.
- We use a mixture of ethanol and water for the Saponification because it prevents the fat from reacting with atmospheric oxygen.
- This mixture is advantageous as it exhibits lower polarity than water, aiding in the dissolution of non-polar fats, thereby facilitating reaction with sodium hydroxide.
Utilize the principle that pH = log { 1 / [H+] }. Designate x as the hydrogen ion concentration of one solution and 100x for the other. The pH of the solution with hydrogen concentration x is pH1 = log {1 / x}. For the solution with 100x concentration, it is pH2 = log {1 / 100x}. Now, you find pH2 - pH1 = log {1/x} - log {1 / 100x}.
By applying the properties of logarithms, you arrive at pH2 - pH1 = log {1/x} - log {1/x} - log {1/100} = - (-2) = 2. Thus, the conclusion is that if one solution contains 100 times more hydrogen ions than another, the difference in pH units between the two solutions is 2<span>.</span>
Vapor pressure refers to the force exerted by vapor or gas molecules above the surface of a liquid. It is inversely related to the concentration of solute particles; an increase in solute concentration results in a decrease in vapor pressure, and vice versa. For (a), it dissociates into two particles. In (b), the total count of particles from dissociation becomes 1 + 2, totaling three. For (c), dissociation yields 1 + 3 for a total of four particles. (d) Since sucrose is a covalent compound, it does not break apart into ions, so it remains as one particle. For (e), dissociation results in 1 + 1, equating to two particles.
Assuming we have a 100g sample, the mass of each element is as follows:
C: 74 g
H: 7.4 g
N: 8.6 g
O: 10 g
Next, we calculate the moles of each by dividing the mass of each element by its molar mass:
C: (74 / 12) = 6.17
H: (7.4 / 1) = 7.4
N: (8.6 / 14) = 0.61
O: (10 / 16) = 0.625
Now, we take the smallest value to determine the ratio:
C: 10
H: 12
N: 1
O: 1
Thus, the empirical formula can be expressed as
C10H12NO
Answer:
0.5 g/mL----- will float
1.0 g/mL---- will float
2.0 g/mL----- will sink
Explanation:
Objects with a density less than or equal to that of water will float due to having a lower mass, while objects with a density exceeding that of water will sink because their mass is greater than that of water. Thus, objects with a density of 0.5 g/mL and 1.0 g/mL will float since they are less dense than water (1 g/mL), whereas an object with a density of 2.0 g/mL will sink.
